If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to
(A) 50° (B) 60° (C) 70° (D) 80°
Solution:
Let's draw a figure as per the question.
The lengths of tangents drawn from an external point to a circle are equal.
A tangent at any point of a circle is perpendicular to the radius at the point of contact.
In ΔOAP and in ΔOBP
OA = OB (radii of the circle are always equal)
AP = BP (length of the tangents)
OP = OP (common)
Therefore, by SSS congruency ΔOAP ≅ ΔOBP
SSS congruence rule: If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.
If two triangles are congruent then their corresponding parts are equal.
Hence,
∠POA = ∠POB
∠OPA = ∠OPB
Therefore, OP is the angle bisector of ∠APB and ∠AOB
Hence, ∠OPA = ∠OPB = 1/2 (∠APB )
= 1/2 × 80°
= 40°
By angle sum property of a triangle,
In ΔOAP
∠A + ∠POA + ∠OPA = 180°
OA ⊥ AP (Theorem 10.1 : The tangent at any point of a circle is perpendicular to the radius through the point of contact.)
Therefore, ∠A = 90°
90° + ∠POA + 40° = 180°
130° + ∠POA = 180°
∠POA = 180° - 130°
∠POA = 50°
Thus, option (A) 50° is the correct answer.
☛ Check: NCERT Solutions Class 10 Maths Chapter 10
Video Solution:
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to (A) 50° (B) 60° (C) 70° (D) 80°
Maths NCERT Solutions Class 10 Chapter 10 Exercise 10.2 Question 3
Summary:
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to 50°.
☛ Related Questions:
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